Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The trace is the sum of the elements on the diagonal of a matrix. Is there a similar operation for the sum of all the elements in a matrix?

share|cite|improve this question
up vote 15 down vote accepted

I don't know if it has a nice name or notation, but for the matrix $\mathbf A$ you could consider the quadratic form $\mathbf e^\top\mathbf A\mathbf e$, where $\mathbf e$ is the column vector whose entries are all $1$'s.

share|cite|improve this answer

Using the sum of all elements does not contain any information about endomorphisms, which is the reason why you will not find such an operation in the literature.

If this is interesting enough, you can get the sum of all squares using the scalar product $$ \phi(A,B) := \mathrm{tr}(A^T B)$$ In fact $\mathrm{tr}(A^T A) = \sum\limits_{i,j=1} a_{i,j}^2$

share|cite|improve this answer
I think the last two sentences are helpful myself. – Geoff Robinson Jul 30 '12 at 13:16

You can certainly consider the sum of all the entries in a square matrix. But what would it be good for?

Mind that square matrices are a way to write explicitly endomorphisms (i.e. linear transformations of a space into itself) so that any quantity you attach to a matrix should be actually say something about the endomorphisms. Trace and determinant remain unchanged if the matrix $A$ is replaced by the matrix $PAP^{-1}$ where $P$ is any invertible matrix. Thus, trace and determinant are numbers that you can attach to the endomorphism represented by $A$.

It wouldn't be the case for the sum of all entries, which does not remain invariant under the said matrix transformation.

share|cite|improve this answer
However, as vanna's answer can be shown to imply, the sum of the squares of all entries remains invariant under orthogonal conjugations. – Qiaochu Yuan Jul 30 '12 at 15:09
@QiaochuYuan : Not surprisingly, since the sum of the squares of the entries is just the square of the norm of $A$ thought as a vector in ${\Bbb R}^{n^2}$. – Andrea Mori Jul 30 '12 at 15:19
I do not think it is completely clear that the Euclidean norm in $\mathbb{R}^{n^2}$ is invariant under conjugation by orthogonal elements, which are defined using the Euclidean norm in $\mathbb{R}^n$. – Qiaochu Yuan Jul 30 '12 at 15:27

The term "grand sum" is commonly used, if only informally, to represent the sum of all elements.

By the way, the grand sum is a very important quantity in the contexts of Markovian transition matrices and other probabilistic applications of linear algebra.

Regards, Scott

share|cite|improve this answer

The max norm:

The max norm is the elementwise norm with $p = \infty$: $$ \|A\|_{\text{max}} = \max \{|a_{ij}|\}. $$ This norm is not sub-multiplicative. $p=\infty$ refers to $\Vert A \Vert_{p} = \left( \sum_{i=1}^m \sum_{j=1}^n |a_{ij}|^p \right)^{1/p}. \, $

If you want something without absolute bars, think of the projection of your matrix on $E$, $\text{tr}\left(E\cdot A\right)$, where $E$ is a matrix full of $1$'s, which is equivalent to calculate the scalar product $\langle e |Ae \rangle$, with $e$ being a vector full of $1$'s, since $|e \rangle \langle e|=E$.

share|cite|improve this answer

I refer you to the article Merikoski: On the trace and the sum of elements of a matrix, Linear Algebra and its applications, Volume 60, August 1984, pp. 177-185.

share|cite|improve this answer
Could you give some description of what that article says? Links are fine, but if the whole answer is essentially a link, it is of little value if the link goes stale. The library reference is also good, but not of much use to someone who doesn't have access to a University Library. – robjohn Jan 31 '13 at 1:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.